Tuesday 11 August 2026

Two decompositions of the payoff

For any real numbers, $max(a,b)=b+max(a-b,0)=a+max(b-a,0)$. Applying this to the terminal payoff two ways:

$$max(S_T,K)=K+(S_T-K{)}^{+}$$

$$max(S_T,K)=S_T+(K-S_T{)}^{+}$$

The first is a zero-coupon bond paying $K$ at $T$ plus a European call struck at $K$. The second is one share held to $T$ plus a European put struck at $K$.

Because these are equalities of payoffs that hold state-by-state at $T$, the law of one price prices each replicating portfolio by summing the prices of its traded components. Let $C$ and $P$ be the call and put prices, and $V$ the price of the contract.

From the bond-plus-call portfolio:

$$V=K{e}^{-rT}+C$$

From the stock-plus-put portfolio, the value today of receiving one share at $T$ (a prepaid forward) is $S_0{e}^{-qT}$, since dividends accruing over $[0,T]$ are forfeited by not holding the share directly:

$$V=S_0{e}^{-qT}+P$$

The step candidates miss is exactly this last one: the stock leg is worth $S_0{e}^{-qT}$, not $S_0$. Discounting the share at the dividend yield is what keeps the two answers consistent.

Why the two prices must agree

Equating the two expressions for $V$:

$$K{e}^{-rT}+C=S_0{e}^{-qT}+P$$

This is precisely put-call parity with a continuous dividend yield. It is not an extra assumption — it is forced by no-arbitrage (the law of one price), because both portfolios have identical payoffs at $T$ and hence identical value at every earlier time. So the two replications give the same $V$ automatically.

Numbers

Discount factors:

$$e^{-rT}=e^{-0.05}=0.951229,e^{-qT}=e^{-0.02}=0.980199$$

Contract price via bond-plus-call:

$$V=100·0.951229+8=95.1229+8=103.1229$$

Put price via parity, $P=C-S_0{e}^{-qT}+K{e}^{-rT}$:

$$P=8-100·0.980199+100·0.951229=8-98.0199+95.1229=5.1030$$

Consistency check with the stock-plus-put portfolio:

$$S_0{e}^{-qT}+P=98.0199+5.1030=103.1229✓$$

Answer: the $max(S_T,K)$ contract is worth $\$103.12$, and the European put struck at $K$ is worth $\$5.10$.

Closing note. The tempting wrong answer for the put uses $S_0$ in place of $S_0{e}^{-qT}$, giving $P=8-100+95.1229=\$3.12$ — an error of about a dollar here, and unbounded as $q$ or $T$ grows. Whenever the underlying pays a yield, replace the spot in every parity/replication relation by the prepaid forward $S_0{e}^{-qT}$.

Setup

For any estimator $T$ of $\mu$, the identity $$𝔼[(T-\mu {)}^2]=(𝔼[T]-\mu {)}^2+\mathrm{V}\mathrm{a}\mathrm{r}(T)$$ holds by adding and subtracting $𝔼[T]$ and noting the cross term $𝔼[(T-𝔼T)(𝔼T-\mu )]$ vanishes because $𝔼T-\mu$ is constant and $𝔼[T-𝔼T]=0$. This is the whole engine of the problem.

Here ${\overline{X}}_n$ has $𝔼[{\overline{X}}_n]=\mu$ and $\mathrm{V}\mathrm{a}\mathrm{r}({\overline{X}}_n)={\sigma }^2/n$, so for ${\widehat{\mu }}_{\lambda }=\lambda {\overline{X}}_n$: $$𝔼[{\widehat{\mu }}_{\lambda }]=\lambda \mu ,\mathrm{V}\mathrm{a}\mathrm{r}({\widehat{\mu }}_{\lambda })={\lambda }^2\frac{{\sigma }^2}{n}.$$

Part 1: the decomposition

The bias is $𝔼[{\widehat{\mu }}_{\lambda }]-\mu =(\lambda -1)\mu$, so $$\mathrm{M}\mathrm{S}\mathrm{E}(\lambda )=\underset{{\text{bias}}^2}{\underbrace{(\lambda -1{)}^2{\mu }^2}}+\underset{\text{variance}}{\underbrace{{\lambda }^2\frac{{\sigma }^2}{n}}}.$$

Shrinking ($\lambda <1$) trades increased squared bias for decreased variance.

Part 2: optimal shrinkage

$\mathrm{M}\mathrm{S}\mathrm{E}(\lambda )$ is a strictly convex quadratic in $\lambda$. Differentiate and set to zero: $$\frac{d}{d\lambda }\mathrm{M}\mathrm{S}\mathrm{E}(\lambda )=2(\lambda -1){\mu }^2+2\lambda \frac{{\sigma }^2}{n}=0.$$

Solving, $${\lambda }^{\backslash }*=\frac{{\mu }^2}{{\mu }^2+{\sigma }^2/n}=\frac{n{\mu }^2}{n{\mu }^2+{\sigma }^2}\in (0,1).$$

To evaluate the minimized risk, use ${\lambda }^{\backslash }*-1=-{\displaystyle \frac{{\sigma }^2/n}{{\mu }^2+{\sigma }^2/n}}$. Then $$({\lambda }^{\backslash }*-1{)}^2{\mu }^2=\frac{({\sigma }^2/n{)}^2{\mu }^2}{({\mu }^2+{\sigma }^2/n{)}^2},({\lambda }^{\backslash }*{)}^2\frac{{\sigma }^2}{n}=\frac{{\mu }^4{\sigma }^2/n}{({\mu }^2+{\sigma }^2/n{)}^2}.$$

Adding, the numerator factors as ${\mu }^2({\sigma }^2/n)({\sigma }^2/n+{\mu }^2)$, cancelling one power of the denominator: $$\mathrm{M}\mathrm{S}\mathrm{E}({\lambda }^{\backslash }*)=\frac{{\mu }^2{\sigma }^2/n}{{\mu }^2+{\sigma }^2/n}=\frac{{\mu }^2{\sigma }^2}{n{\mu }^2+{\sigma }^2}.$$

A cleaner way to see the same value: $\mathrm{M}\mathrm{S}\mathrm{E}({\lambda }^{\backslash }*)=({\sigma }^2/n){\lambda }^{\backslash }*$, since at the optimum $({\lambda }^{\backslash }*-1){\mu }^2=-{\lambda }^{\backslash }*{\sigma }^2/n$ makes bias${}^2=(1-{\lambda }^{\backslash }*){\lambda }^{\backslash }*{\sigma }^2/n$ and variance$=({\lambda }^{\backslash }*{)}^2{\sigma }^2/n$, summing to ${\lambda }^{\backslash }*{\sigma }^2/n$.

Part 3: comparison and the catch

For the unbiased estimator $\lambda =1$: $\mathrm{M}\mathrm{S}\mathrm{E}(1)={\sigma }^2/n$. Since ${\lambda }^{\backslash }*<1$, $$\mathrm{M}\mathrm{S}\mathrm{E}({\lambda }^{\backslash }*)={\lambda }^{\backslash }*\frac{{\sigma }^2}{n}<\frac{{\sigma }^2}{n}=\mathrm{M}\mathrm{S}\mathrm{E}(1)$$ strictly, for every $\mu \ne 0$. So the unbiased estimator is inadmissible under squared error in this one-parameter family: a biased estimator dominates it.

The obstruction: ${\lambda }^{\backslash }*$ depends on the unknown $\mu$ (through ${\mu }^2$), so it is not an estimator one can compute. This is exactly the tension that James-Stein shrinkage resolves in dimension $\ge 3$ by estimating the shrinkage factor from the data, achieving uniform dominance without knowing $\mu$.

Answers. 1. $\mathrm{M}\mathrm{S}\mathrm{E}(\lambda )=(\lambda -1{)}^2{\mu }^2+{\lambda }^2{\sigma }^2/n$. 2. ${\lambda }^{\backslash }*={\displaystyle \frac{n{\mu }^2}{n{\mu }^2+{\sigma }^2}}$, with $\mathrm{M}\mathrm{S}\mathrm{E}({\lambda }^{\backslash }*)={\displaystyle \frac{{\mu }^2{\sigma }^2}{n{\mu }^2+{\sigma }^2}}$. 3. $\mathrm{M}\mathrm{S}\mathrm{E}({\lambda }^{\backslash }*)={\lambda }^{\backslash }*{\sigma }^2/n<{\sigma }^2/n=\mathrm{M}\mathrm{S}\mathrm{E}(1)$ for all $\mu \ne 0$; but ${\lambda }^{\backslash }*$ is not usable since it depends on the unknown $\mu$.